Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{-9z - 72}{z - 7} \div \dfrac{z + 8}{z^2 + 3z - 70} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-9z - 72}{z - 7} \times \dfrac{z^2 + 3z - 70}{z + 8} $ First factor the quadratic. $q = \dfrac{-9z - 72}{z - 7} \times \dfrac{(z - 7)(z + 10)}{z + 8} $ Then factor out any other terms. $q = \dfrac{-9(z + 8)}{z - 7} \times \dfrac{(z - 7)(z + 10)}{z + 8} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -9(z + 8) \times (z - 7)(z + 10) } { (z - 7) \times (z + 8) } $ $q = \dfrac{ -9(z + 8)(z - 7)(z + 10)}{ (z - 7)(z + 8)} $ Notice that $(z + 8)$ and $(z - 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -9\cancel{(z + 8)}(z - 7)(z + 10)}{ \cancel{(z - 7)}(z + 8)} $ We are dividing by $z - 7$ , so $z - 7 \neq 0$ Therefore, $z \neq 7$ $q = \dfrac{ -9\cancel{(z + 8)}\cancel{(z - 7)}(z + 10)}{ \cancel{(z - 7)}\cancel{(z + 8)}} $ We are dividing by $z + 8$ , so $z + 8 \neq 0$ Therefore, $z \neq -8$ $q = -9(z + 10) ; \space z \neq 7 ; \space z \neq -8 $